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  1. #1
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    Smooth

    Let  f: \mathbb{R} \to \mathbb{R} be given. Assume that the square and cube of  f are smooth. Is  f smooth? That is , if  f \cdot f \in C^{\infty} and  f \cdot f \cdot f \in C^{\infty} , does it follow that  f \in C^{\infty} ?
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  2. #2
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    Or maybe we can put it this way: if  f^{n} \in C^{\infty} and  f^{n-1} \in C^{\infty} does it follow that  f \in C^{\infty} ?
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  3. #3
    Super Member Rebesques's Avatar
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    Or maybe we can put it this way: if
    <br />
f^{n} \in C^{\infty}<br />
and <br />
f^{n-1} \in C^{\infty}<br />
does it follow that <br />
f \in C^{\infty}<br />
?

    Not in general. You need the extra assumption that f^n,f^{n-1} are never zero.

    The proof is easy: Choose any natural number m. Since gcd(n,n-1)=1, there are integers q, \ p such that pn+q(n-1)=1. So m=pmn+qm(n-1); Thus f^m=(f^n)^{mp}(f^{n-1})^{qm} is differentiable as a product of differentiable functions. Note that the exponents mp, \ mq may be negative, and so the assumption f^n, f^{n-1}\neq0 is essential.
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  4. #4
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    I have another way to show this.

    If f^3\not = 0 for any point and f^3\in C^{\infty} then (f^3)^{1/3} \in C^{\infty} thus f\in C^{\infty}. Because the composition of infinitely differenciable functions is infinitely differenciable. Note, that x^{1/3} is not differenciable at zero, which is why we need the condition that the function is never zero. However, I cannot come up with a conterexample.


    This does not work with f^2.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Rebesques View Post
    Not in general. You need the extra assumption that f^n,f^{n-1} are never zero.

    The proof is easy: Choose any natural number m. Since gcd(n,n-1)=1, there are integers q, \ p such that pn+q(n-1)=1. So m=pmn+qm(n-1); Thus f^m=(f^n)^{mp}(f^{n-1})^{qm} is differentiable as a product of differentiable functions. Note that the exponents mp, \ mq may be negative, and so the assumption f^n, f^{n-1}\neq0 is essential.
    Ah! So my thought that we may construct the function f as \frac{f^3}{f^2} is at least not out of the ballpark. (This, of course, gives the same limitation that f \neq 0 at any point in its domain.) But this can't be the whole story because if we have
    f^2(x) = (x - 1)^4
    and
    f^3(x) = (x - 1)^6

    the function f(x) = (x - 1)^2 is, in fact, C^{\infty}.

    How might we get around this? (Or am I barking up the wrong tree again? )

    -Dan
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  6. #6
    Super Member Rebesques's Avatar
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    Topsq, you have chosen a polynomial

    Powers of that are always differentiable!



    ps. What? 300 posts!! The Hacker must be feeling my breath on his back... :P
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Rebesques View Post
    Topsq, you have chosen a polynomial

    Powers of that are always differentiable!



    ps. What? 300 posts!! The Hacker must be feeling my breath on his back... :P
    But my f^3 and f^2 are C^{\infty} and there is a point in the domain of both that are 0. According to you and TPH my f should not be C^{\infty}. Or should we all be saying that we can't guarantee that such a C^{\infty} f is possible under these conditions?

    -Dan
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  8. #8
    Super Member Rebesques's Avatar
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    According to you and TPH my f should not be...

    Alright you win, add "polynomials and no-denominator ever functions out"
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